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Abs max/min

  1. Jul 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the absolute max and min values of f on the set D.

    f(x,y)=4xy^3 - (x^2)(y^2) - xy^3
    D is the closed triangular region in the xy-plane with vertices (0,0) (0,6) and (6,0).

    3. The attempt at a solution
    I found my two critical points to be (1,2) and (2,0). Then I tried to evaluate the boundary points:
    1) 0<x<6, y=0
    2) 0<y<6, x=0
    3) (6-y, y) because the third boundary line is y= -x+6

    I don't know how to solve for the last boundary line though. I plugged in x=6-y in the original equation, got the expression (2y^3)-(12y^2). Do I just plug in numbers now?

    My book gets (2,4) for the absolute min, which is a point on this 3rd boundary line. I just don't see how to come up with the point, though.
  2. jcsd
  3. Jul 20, 2008 #2


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    Staff Emeritus
    Science Advisor

    Did you write this correctly? Why not just 3xy3- x2y2?

    f(x, 0)= 0 so that's easy.

    f(0,y)= 0 so that's easy.

    If what you wrote before is correct, f(6-y,y) will be a fourth degree polynomial, not a cubic. In any case, what every you get as a function of y alone, with y between 0 and 6, you differentiate and set equal to 0 to find the max or min.

    Last edited: Jul 21, 2008
  4. Jul 20, 2008 #3
    Sorry! It is a typo. It should read:
    Find the absolute max and min values of f on the set D.

    f(x,y)=4xy^2 - (x^2)(y^2) - xy^3

    So, if the other values of f(0,y) and f(x,0) did not give values of 0, would I also need to differentiate their functions and find the critical points there as well?
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